Calculus Tutorial: Chain, Product, and Quotient Rule

Alexa W. and Nathan H.

Chain Rule:

The Chain Rule enables our blossoming undergrads to take the derivative of composite functions, such as $f(g(x))$. We do this by taking the derivative of $f(z)$ where $z=g(x)$, then multiplying $f’(z)$ by $g’(x)$. So the formula would be:

$$\frac{d}{dx} f(g(x)) = f'(g(x)) g'(x)$$

Here is an example. Let’s use the Chain Rule to take the derivative of this function:
$$p(x) = (x^2 + 3x + 2)^7$$

First we write the function, as shown above. Next, we can isolate the two different functions we have to take the derivatives of. $$p(z) = z^7$$
And $$z= m(x) = x^2 + 3x + 2$$
So now that we have the two functions, we can apply our rule:
$$p'(x) = 7 \left( m(x) \right)^6 \cdot \left(2x + 3\right)$$ $$p'(x) = 7 \left( x^2 + 3x + 2 \right)^6 \cdot \left(2x + 3\right)$$

This is the derivative. We can simplify it if we want to and we would get:
$$p'(x) = \left( x^2 + 3x + 2 \right)^6 \cdot \left(14x + 21\right)$$
You now know the Chain Rule. Hurray!

Product Rule:

This is another important thing to remember for derivatives. This rule can be used when you have to find the derivative of a function in the form of $f(x)g(x)$. The formula for this rule is: $$\frac{d}{dx}f(x)g(x) = f'(x) g(x) + g'(x) f(x)$$
Let’s try an example. Here’s our function: $$f(x) = (x^2 + 1)(3x^3 - 7)$$
Then we can take our rule and apply it:
$$f'(x) = \left( \frac{d}{dx}(x^2 + 1) \right) (3x^2 - 7) + \left( \frac{d}{dx}(3x^3 - 7) \right) (x^2 +1)$$
$$f'(x) = (2x)(3x^3 -7) + 6x^2 (x^2 + 1)$$
And if we want, we can simplify this function to get:
$$f'(x) = 12x^4 + 6x^2 - 14x$$
You have now learned the product rule.


Keep going you're almost done!!!

Quotient Rule:

The quotient rule can be applied when you have to take the derivative of the quotient of two functions. For instance, you can use the quotient rule to handle the derivatives of rational functions. These are functions that have a polynomial in the denominator and in the numerator. They look like this:
$$f(x) = h(x)/g(x)$$
They’re pretty great. But they’re not fun to derive. So we came up with this amazing rule to help us: $$\frac{d}{dx} f(x) = \frac{g(x) h'(x) - h(x) g'(x)}{g(x)^2}$$

Now that we have this amazing rule, let’s try it out. Here’s our function:
$$f(x) = \frac{7x^3 + 3x}{3x + 1}$$
Let’s apply our rule:
$$f'(x) = \frac{(3x+1)(21x^2 + 3) - (7x^3 + 3x)(3)}{(3x + 1)^2}$$


This is our derivative! We don’t need to simplify. But we can. But we won’t. It’s not needed.

Practice Problems: So now that we know how to take the derivatives of these functions, here’s some good practice problems that you should do: $$z(q) = 16(q^4 + 26)^3$$
$$g(p) = (p^3 + 6) (p - 12)$$
$$w(x) = \frac{81x + 1187924}{67x^{896} + 12}$$



Have fun with those terrible numbers. Also, here’s a really ugly one you can try if you’re feeling up for it:
$$g(t) = \frac{(t+ 17) \cos\left( e^{\cos( e^{\cos(e^{t} ) })} \right)}{1 + t^2}$$

Good luck!